Term n of an arithmetic series given by

Tn= a + (n-1)d

where a = first term and d = common difference

d is given = 3

T5 = a + (5-1)3 = a + 12 = 14

so a = 2

Sn the summation of the arithmetic series is given by

Sn= n/2[2a +(n-1)d]

so in this case Sn= n/2[4+3(n-1)]

and S2n= 2n/2[4+3(2n-1)]

S2n - Sn reduces to n/2(9n+1)